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Aliun [14]
3 years ago
13

The length of rectangle is represented by x and the width by y. The square of the diagonal of the rectangle is equal to the sum

of the squares of the length and the width. If the length is 25 meters and the diagonal is 45 meters, which quadratic equation could be used to determine the width of the rectangle?
Mathematics
2 answers:
maksim [4K]3 years ago
8 0
To solve this problem you must apply the proccedure shown below:

 1- You have the following information given in the problem above:

 - <span>The square of the diagonal of the rectangle is equal to the sum of the squares of the length and the width.
 - The length is 25 meters and the diagonal is 45 meters.

 2- Therefore, you have:

 x: diagonal of the rectangle.

 x^2=l^2+w^2

 3- You have:

 w^2+l^2-x^2=0
 w^2+(25)^2-(45)^2=0
 w^2-1400=0
 w=37.41

 The answer is: </span>w^2-1400=0
Artyom0805 [142]3 years ago
8 0

Answer:

B) 252 + y^2 = 452

Step-by-step explanation:

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Answer: I got -14x-13

But it sounds wrong but I have a picture

Step-by-step explanation:

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4 years ago
Henry know that the circumference of a circle is 18 pie inches. What is the area of the circle?
Leto [7]

Answer:

B: 81π inches

Step-by-step explanation:

The formula for the circumference of a circle is C = πd and the formula for the area of a circle is A = πr²

  • The formula for the circumference of a circle can also be written as C = 2πr since the diameter is equal to double the radius (d = 2r)
  • So we need to solve for r since that is the variable we are missing in order to solve for the area

Plugging in our known values to the circumference formula we have:

  • 18π = 2πr
  • Then dividing both sides by 2π we have 9 = r

Now that we have solved for the radius, we can plug in this value to the area formula:

  • A = π(9)² = 81π in
3 0
4 years ago
C=1/21.22.23+1/22.23.24+................+1/200.201.202<br><br> . = là dấu nhân
Aneli [31]

It looks like you have to find the value of the sum,

C = \displaystyle \frac1{21\times22\times23} + \frac1{22\times23\times24} + \cdots + \frac1{200\times201\times202}

so that the <em>n</em>-th term in the sum is

\dfrac1{(21+(n-1))\times(21+n)\times(21+(n+1))} = \dfrac1{(n+20)(n+21)(n+22)}

for 1 ≤ <em>n</em> ≤ 180.

We can then write the sum as

\displaystyle C = \sum_{n=1}^{180} \frac1{(n+20)(n+21)(n+22)}

Break up the summand into partial fractions:

\dfrac1{(n+20)(n+21)(n+22)} = \dfrac a{n+20} + \dfrac b{n+21} + \dfrac c{n+22}

Combine the fractions into one with a common denominator and set the numerators equal to one another:

1 = a(n+21)(n+22) + b(n+20)(n+22) + c(n+20)(n+21)

Expand the right side and collect terms with the same power of <em>n</em> :

1 = a(n^2+43n+462)+b(n^2+42n+440) + c(n^2+41n + 420) \\\\ 1 = (a+b+c)n^2 + (43a+42b+41c)n + 462a+440b+420c

Then

<em>a</em> + <em>b</em> + <em>c</em> = 0

43<em>a</em> + 42<em>b</em> + 41<em>c</em> = 0

462<em>a</em> + 440<em>b</em> + 420<em>c</em> = 1

==>   <em>a</em> = 1/2, <em>b</em> = -1, <em>c</em> = 1/2

Now our sum is

\displaystyle C = \sum_{n=1}^{180} \left(\frac1{2(n+20)}-\frac1{n+21}+\frac1{2(n+22)}\right)

which is a telescoping sum. If we write out the first and last few terms, we have

<em>C</em> = 1/(2×21) - 1/22 <u>+ 1/(2×23)</u>

… … + 1/(2×22) - 1/23 <u>+ 1/(2×24)</u>

… … <u>+ 1/(2×23)</u> - 1/24 <u>+ 1/(2×25)</u>

… … <u>+ 1/(2×24)</u> - 1/25 <u>+ 1/(2×26)</u>

… … + … - … + …

… … <u>+ 1/(2×198)</u> - 1/199 <u>+ 1/(2×200)</u>

… … <u>+ 1/(2×199)</u> - 1/200 + 1/(2×201)

… … <u>+ 1/(2×200)</u> - 1/201 + 1/(2×202)

Notice the diagonal pattern of underlined and bolded terms that add up to zero (e.g. 1/(2×23) - 1/23 + 1/(2×23) = 1/23 - 1/23 = 0). So, like a telescope, the sum collapses down to a simple sum of just six terms,

<em>C</em> = 1/(2×21) - 1/22 + 1/(2×22) + 1/(2×201) - 1/201 + 1/(2×202)

which we simplify further to

<em>C</em> = 1/42 - 1/44 - 1/402 + 1/404

<em>C</em> = 1,115/1,042,118 ≈ 0.00106994

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ludmilkaskok [199]

Answer:

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Step-by-step explanation:

A graphing calculator can plot this function for you.

If you divide numerator and denominator by 25, you get ...

  r = 1/(0.4 -sin(θ))

This will have asymptotes at θ = arcsin(0.4). At those values of θ, the value of r will change sign, so the graph cannot be a parabola or ellipse.

The graph of the hyperbola is the appropriate choice.

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Zolol [24]
23.4 ounces of Oreos
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